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Show that if $X$ and $Y$ are independent $N(0, 1)$-distributed random variables, then $X/Y ∈ C(0, 1)$. C is cauchy distribution..

First trial: 1) Transformation

Let $Y=V,U=X/V,=>X=UV$;$|J|=v$

$f(u,v)=f_x(uv)f_y(v)|J|=1/2\pi ve^{-\frac{(uv)^2}{2}}e^{-\frac{v^2}2}$

To get $f_u(u) =\int_{-∞}^{+∞}1/2\pi ve^{-\frac{(uv)^2}{2}}e^{-\frac{v^2}2}=0$

Second trial: 2) CDF

$F(U\le u)=F(X \le uY)=\int_{-∞}^{+∞}F(X \le uy)f_y(y)dy=1/2\pi\int_{-∞}^{+∞} (\int_{-∞}^{uy}e^{-\frac{(uy)^2}{2}}du)e^{-\frac{y^2}2}dy=?$

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I carry on your second trial: $\frac{X}Y\leq u $

Now we discriminate between two cases $Y>0$ and $Y<0$. Multiplying the equation above by $Y$.

$P(\frac{X}Y\leq u)=(X\leq uY| Y>0)+P(X\geq uY| Y<0)$

$=\int_{0}^{\infty}\int_{0}^{uy} f(x,y) \,dx \, dy +\int_{-\infty}^{0}\int_{uy}^{0} \,dx \, dy$

Then we differentiate w.r.t. u to get the pdf. This can be done by using the Leibniz rule.

$p(u)=\int_{0}^{\infty} y\cdot f(uy,y) \, dy- \int_{-\infty}^{0} y\cdot f(uy,y) \, dy$

$=2\int_{0}^{\infty} y\cdot f(uy,y) \, dy$

$=2\cdot\frac{1}{2\pi\sigma_x\sigma_y}\int_{0}^{\infty} y\cdot \exp\left( -\left(\frac{y^2}{2\sigma_x^2}+\frac{u^2y^2}{2\sigma_y^2}\right)\right) \, dy$

$=\frac{1}{\pi\sigma_x\sigma_y}\int_{0}^{\infty} y\cdot \exp\left( -y^2\left(\frac{1}{2\sigma_x^2}+\frac{u^2}{2\sigma_y^2}\right)\right) \, dy$

Let $c=\frac{1}{2\sigma_x^2}+\frac{u^2}{2\sigma_y^2}$. The integral becomes equal to

$\int_{0}^{\infty} y\cdot \exp\left( -cy^2\right) \, dy$, which is $\frac1{2c}$. Thus we have

$p(u)=\frac{1}{\pi\sigma_x\sigma_y}\cdot \frac{1}{2\cdot(\frac{1}{ 2\sigma_x^2}+\frac{u^2}{2\sigma_y^2})}=\frac{1}{\pi\sigma_x\sigma_y}\cdot \frac{1}{\frac{1}{ \sigma_x^2}+\frac{u^2}{\sigma_y^2}}$

$=\boxed{\large{\frac1{\pi}\frac{\frac{\sigma_y}{\sigma_x}}{u^2+\left(\frac{\sigma_y}{\sigma_x}\right)^2}}}$

This is the pdf of the cauchy distribution with $\gamma=\frac{\sigma_y}{\sigma_x}$ and $x_0=0$.

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